P4767 IOI2000 邮局

有 \(n\) 个村庄，给出了他们的横坐标。现在要在这些村庄中的一些村庄安放 \(p\) 个邮局，要求算每个村庄和最近的邮局之间所有距离的最小可能的总和。

\(50\%:\sim70\%\)

悄咪咪告诉你，开氧可以跑\(90pts!!!!!!!!!!\) 这还打啥正解

\(f[i][j]\)表示前\(i\)个村庄放\(j\)个邮局最小距离和

\(w[i][j]\)表示村庄区间\([i,j]\)内放\(1\)个邮局的距离和

\(w\)可以预处理出来，放置一个邮局，邮局位置总是在中位数处

\(w[l][r]=w[l][r-1]+X[r]-X[\frac{l+r}2]\) \(V^2\)递推

\(f[i][j]=\min\{f[k][j-1]+w[k+1][j])\}~~k\in[0,i)\)

复杂度\(O(PV^2)\)

const int N = 3005;
int V,P,x[N],w[N][N],f[N][305];
inline int min(int a,int b){return a < b ? a : b;}
void init(){
	for(re int i = 1;i <= V;++i) x[i] = read();
	for(re int i = 1;i <= V;++i)
		for(re int j = i+1;j <= V;++j)
			w[i][j] = w[i][j-1] + x[j] - x[i+j>>1];
}
int main(){
	V = read(); P = read();
	init();
	memset(f,0x3f,sizeof(f));
	f[0][0] = 0;
	for(re int j = 1;j <= P;++j)
		for(re int i = 1;i <= V;++i)
			for(re int k = 0;k < i;++k)
				f[i][j] = min(f[k][j-1] + w[k+1][i],f[i][j]);
	printf("%d",f[V][P]);
}

\(100\%:\)

\(w\)其实是满足四边形不等式的

由递推式可得\(w[l][r+1]-w[l][r]=X[r+1]-X[\frac{l+r+1}2]~~①\)

\[w[l+1][r+1]-w[l+1][r]=X[r+1]-X[\frac{l+r+2}2]~~②\\ ①-②\\ =X[\frac{l+r+2}2]-X[\frac{l+r+1}2]\\ \]

\(X\)递增，则\(\large X[\frac{l+r+2}2]-X[\frac{l+r+1}2]\ge0\)

即

\[w[l][r+1]-w[l][r]-(w[l+1][r+1]-w[l+1][r])\ge0\\ w[l][r+1]-w[l][r]-w[l+1][r+1]+w[l+1][r]\ge0\\ w[l][r+1]+w[l+1][r]\ge w[l][r]+w[l+1][r+1] \]

显然\(w\)满足四边形不等式，且对于\(a\le b\le c\le d\)，\(w[a][d]\ge w[b][c]\)，\(dp\)满足四边形不等式

所以对于\(f[i][j]\)决策，\(f[i][j-1]\le f[i][j]\le f[i+1][j]\)

令\(s[i][j]\)为\(i\)之前建\(j\)个邮局的最优决策

所以转移\(f[i][j]\)时，从\([s[i][j-1],s[i+1][j]]\)中找最优决策

因为要用到\(f[i+1][j]\)，所以倒叙枚举\(i\)

\(O(PV)\)

int V,P,x[N],w[N][N],f[N][305],s[N][305];
void init(){
	for(int i = 1;i <= V;++i) x[i] = read();
	for(int i = 1;i <= V;++i)
		for(int j = i+1;j <= V;++j)
			w[i][j] = w[i][j-1] + x[j] - x[i+j>>1];
}
int main(){
	V = read(); P = read();
	init();
	memset(f,0x3f,sizeof(f));
	f[0][0] = 0;
	for(int j = 1;j <= P;++j){
		s[V + 1][j] = V;
		for(int i = V;i >= 1;--i){ 
			int minn = 0x3f3f3f3f,minid;
			for(int k = s[i][j-1];k <= s[i+1][j];++k){
				if(f[k][j-1] + w[k+1][i] < minn){
					minn = f[k][j-1] + w[k+1][i];
					minid = k;
				}
			}
			f[i][j] = minn; s[i][j] = minid;	
		} 
	}
	printf("%d",f[V][P]);
}